Leecode 0012. Integer to Roman
12. Integer to Roman
题目
Roman numerals are represented by seven different symbols: I
, V
, X
, L
, C
, D
and M
.
1 | Symbol Value |
For example, 2
is written as II
in Roman numeral, just two one's added together. 12
is written as XII
, which is simply X + II
. The number 27
is written as XXVII
, which is XX + V + II
.
Roman numerals are usually written largest to smallest from left to right. However, the numeral for four is not IIII
. Instead, the number four is written as IV
. Because the one is before the five we subtract it making four. The same principle applies to the number nine, which is written as IX
. There are six instances where subtraction is used:
I
can be placed beforeV
(5) andX
(10) to make 4 and 9.X
can be placed beforeL
(50) andC
(100) to make 40 and 90.C
can be placed beforeD
(500) andM
(1000) to make 400 and 900.
Given an integer, convert it to a roman numeral.
Example 1:
1 | Input: num = 3 |
Example 2:
1 | Input: num = 4 |
Example 3:
1 | Input: num = 9 |
Example 4:
1 | Input: num = 58 |
Example 5:
1 | Input: num = 1994 |
Constraints:
1 <= num <= 3999
题目大意
通常情况下,罗马数字中小的数字在大的数字的右边。但也存在特例,例如 4 不写做 IIII,而是 IV。数字 1 在数字 5 的左边,所表示的数等于大数 5 减小数 1 得到的数值 4 。同样地,数字 9 表示为 IX。这个特殊的规则只适用于以下六种情况:
- I 可以放在 V (5) 和 X (10) 的左边,来表示 4 和 9。
- X 可以放在 L (50) 和 C (100) 的左边,来表示 40 和 90。
- C 可以放在 D (500) 和 M (1000) 的左边,来表示 400 和 900。
给定一个整数,将其转为罗马数字。输入确保在 1 到 3999 的范围内。
解题思路
贪心算法,利用罗马数字的符号与对应数值的映射关系,从最大的数值开始逐步递减,直到将整数减为 0:
建立罗马数字符号与对应数值的映射表,按从大到小排序
遍历映射表,对于每个数值:
若当前整数大于等于该数值,将对应的符号添加到结果中
减去该数值,重复上述操作,直到当前整数小于该数值
继续处理下一个较小的数值,直到整数变为 0
代码
1 | #include <string> |